Re: The quadratic Problem

I would do this a completely different way. If we use x and y rather than a and b (it is still exactly the same problem, of course), we can think of "a^2+ b^2= 1" (x^2+ y^2= 1) as the equation of a circle with center at (0, 0) and radius 1. "a+ b= c" or x+ y= c, for c a fixed number, is a line sloping down to the left at a 45 degree angle. The smallest and largest possible values of c are when that line is tangent to the circle which happen at the endpoints of the diameter perpendicular to the line: that is the line y= x which crosses the circle $\displaystyle x^2+ y^2= 1$ when $\displaystyle x^2+ x^2= 2x^2= 1$ so that $\displaystyle x= y= \pm\sqrt{2}/2$. Taking $\displaystyle x= a= b= \sqrt{2}/2$, $\displaystyle a+ b= \sqrt{2}/2+\sqrt{2}/2= \sqrt{2}$. But taking $\displaystyle x= a= b= -\sqrt{2}/2$ we get $\displaystyle a+ b= -\sqrt{2}$.